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Posted: September 22nd, 2023
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Answer the following questions and upload your work into Canvas. Be sure to write your name on your answer sheet, number your answers, show your work, organize your work in a logical and readable order, and upload a single submission (if you have multiple pages of work, take pictures of your work and copy/paste/drop into a single document).
Use VassarStats, with the accompanying recorded demo, to answer the questions below.
Find p from z:
1. Find p > z = 1.28
2. Find p < z = –.37
3. Find p between z = ±.93
Find X from p:
For questions 4 and 5, use the following population parameters to find p from X: μ = 28.29, σ = 3.87.
4. What is the raw score (X) that corresponds to p = .309 in the lower tail of the distribution?
5. What are the raw scores (X’s) that corresponds to p = .92 in the two tails of the distribution combined?
Find z from p:
For questions 6 and 7, use the mean and standard deviation of a normal z-distribution to find p from z.
6. What is the z-score that corresponds to p = .115 in the upper tail of the distribution?
7. What are the z-scores that corresponds to p = .050 in the two tails of the distribution combined?
Find p > z = 1.28
Based on the standard normal distribution table, when z = 1.28, the corresponding p-value is 0.1. Therefore, p > 1.28 is 0.1.
Find p < z = –.37
When z = -0.37, the corresponding p-value is 0.355. Therefore, p < -0.37 is 0.355.
Find p between z = ±.93
When z = 0.93, p = 0.175 in the upper tail. When z = -0.93, p = 0.175 in the lower tail. Therefore, the total p between ±0.93 is 0.175 + 0.175 = 0.35.
What is the raw score (X) that corresponds to p = .309 in the lower tail of the distribution?
Given: μ = 28.29, σ = 3.87
To find the raw score X corresponding to a p-value in the lower tail, we use the inverse normal cumulative distribution function norminv(p, mean, standard deviation)
norminv(0.309, 28.29, 3.87) = 25.5
Therefore, the raw score X corresponding to p = 0.309 in the lower tail is 25.5.
What are the raw scores (X’s) that corresponds to p = .92 in the two tails of the distribution combined?
To find the scores in the two tails corresponding to a combined p-value of 0.92, we calculate 0.92/2 = 0.46 for the p-value of each tail
norminv(0.46, 28.29, 3.87) = 31.5 for the upper tail
norminv(1-0.46, 28.29, 3.87) = 24.9 for the lower tail
Therefore, the raw scores corresponding to a combined p=0.92 are 31.5 and 24.9.
What is the z-score that corresponds to p = .115 in the upper tail of the distribution?
When p = 0.115, the z-score in the upper tail from the standard normal distribution table is 1.28.
What are the z-scores that corresponds to p = .050 in the two tails of the distribution combined?
To find the z-scores corresponding to a combined p-value of 0.05 in the two tails, we calculate:
0.05/2 = 0.025 for each tail
From the standard normal distribution table, z = 1.96 for the upper tail
z = -1.96 for the lower tail
Therefore, the z-scores corresponding to a combined p=0.05 are 1.96 and -1.96.
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